这是杜兰大学计量经济学ECON 3230 Econometrics课程。计算工具是我们日常生活的重要组成部分。软件是尖端科学发现、热门娱乐和当今快节奏市场背后的驱动力。本课程介绍用于开发其中一些工具的技术、想法和解决问题的方法。在高层次上，我们专注于发展“计算思维”，这是使用抽象来设计和实现算法和软件来解决我们日常生活中许多不同领域（例如网络、社交媒体和计算机）中出现的问题的实践。仅举几例，科学计算。在实践层面，学生将设计、实现、测试和记录他们的程序，以学习入门编程概念，例如：数据类型和数据结构（例如列表、字典、树）；编程技术（使用函数的模块化设计、递归、面向对象编程）；通过理论估计、分析和计时进行性能分析。本课程的大部分作业都是编程作业，旨在教会学生用高效、优雅的代码表达自己的想法；无需具备任何编程经验即可加入该课程并取得成功。讲座时间致力于介绍新材料、讨论、个人和小组活动。实验时间用于编程练习。本课程的大部分作业都是编程作业，旨在教会学生用高效、优雅的代码表达自己的想法；无需具备任何编程经验即可加入该课程并取得成功。讲座时间致力于介绍新材料、讨论、个人和小组活动。实验时间用于编程练习。本课程的大部分作业都是编程作业，旨在教会学生用高效、优雅的代码表达自己的想法；无需具备任何编程经验即可加入该课程并取得成功。讲座时间致力于介绍新材料、讨论、个人和小组活动。实验时间用于编程练习。
Using the factorization theorem for sufficient statistics, show that in a $n$ times repeated Bernoulli experiment ( $n$ is known), the number of successes is a sufficient statistic for the success probability $p$.
a. Here is a formulation of the factorization theorem: Given a family of discrete probability measures $\operatorname{Pr}_\theta$ depending on a parameter $\theta$. The statistic $t$ is sufficient for
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parameter $\theta$ iff there exists a function of two variables $g: \mathbb{R} \times \Theta \rightarrow \mathbb{R},(t, \theta) \mapsto g(t ; \theta)$, and a function of one variable $h: U \rightarrow \mathbb{R}, \omega \mapsto h(\omega)$ so that for all $\omega \in U$
$$
\operatorname{Pr}_\theta[{\omega}]=g(t(\omega), \theta) \cdot h(\omega) .
$$
Before you apply this, ask yourself: what is $\omega$ ?
Answer. This is very simple: the probability of every elementary event depends on this element only through the random variable $t: U \rightarrow N$, which is the number of successes. $\operatorname{Pr}[{\omega}]=$ $p^{t(\omega)}(1-p)^{n-t(\omega)}$. Therefore $g(k ; p)=p^k(1-p)^{n-k}$ and $h(\omega)=1$ does the trick. One can also say: the probability of one element $\omega$ is the probability of $t(\omega)$ successes divided by $\left(\begin{array}{c}n \ t(\omega)\end{array}\right)$. This gives another easy-to-understand factorization.
Show that the Binomial distribution
$\quad p_x(k)=\operatorname{Pr}[x=k]=\left(\begin{array}{l}n \ k\end{array}\right) p^k(1-p)^{(n-k)} \quad k=0,1,2, \ldots, n$
is a member of the exponential family. Compute the canonical parameter $\theta$ and the function $b(\theta)$.
Rewrite as
$$
p_x(k)=\left(\begin{array}{l}
n \
k
\end{array}\right)\left(\frac{p}{1-p}\right)^k(1-p)^n=\exp \left(k \ln \left(\frac{p}{1-p}\right)+n \ln (1-p)+\ln \left(\begin{array}{l}
n \
k
\end{array}\right)\right)
$$
therefore $\theta=\ln \left(\frac{p}{1-p}\right)$. To compute $b(\theta)$ you have to express $n \ln (1-p)$ as a function of $\theta$ and then reverse the sign. The following steps are involved: $\exp \theta=\frac{p}{1-p}=\frac{1}{1-p}-1 ; 1+\exp \theta=\frac{1}{1-p}$; $\ln (1+\exp \theta)=-\ln (1-p) ;$ therefore $b(\theta)=n \ln (1+\exp \theta)$.
Show that the Poisson distribution with $t=1$, i.e.,
$$
\operatorname{Pr}[x=k]=\frac{\lambda^k}{k !} e^{-\lambda} \quad \text { for } k=0,1, \ldots
$$
is a member of the exponential family. Compute the canonical parameter $\theta$ and the function $b(\theta)$.
ANSWER. The probability mass function can be written as
$$
\operatorname{Pr}[x=k]=\frac{e^{k \ln \lambda}}{k !} e^{-\lambda}=\exp (k \ln \lambda-\lambda-\ln k !) \quad \text { for } k=0,1, \ldots
$$
This is (6.2.3) for the Poisson distribution, where the values of the random variable are called $k$ instead of $x$, and $\theta=\ln \lambda$. Substituting $\lambda=\exp (\theta)$ in (6.2.7) gives
$$
\operatorname{Pr}[x=k]=\exp (k \theta-\exp (\theta)-\ln k !) \quad \text { for } k=0,1, \ldots .
$$
from which one sees $b(\theta)=\exp (\theta)$.
The one-parameter exponential family can be generalized by the inclusion of a scale parameter $\phi$ in the distribution. This gives the exponential dispersion family, see [MN89, p. 28]: Each observation has the density function
$$
f_y(y ; \theta, \phi)=\exp \left(\frac{y \theta-b(\theta)}{a(\phi)}+c(y, \phi)\right)
$$
